Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2023 (A15757W1)

Source PDF: mmathphys.web.ox.ac.uk/sitefiles/advanced-quantum-theory.pdf

Assume $Q\ge 2$ and $L\ge 2$; the trivial exceptions are described below. In this periodic Potts chain, every satisfied bond contributes $-J$ to the energy, and no bond can contribute less. All $L$ bonds can be satisfied simultaneously by choosing the same state at every site. Conversely, satisfying every bond forces this uniform configuration. Therefore the ground state energy and degeneracy factor areThe $Q$ choices are distinct configurations on the labelled lattice: a global relabelling of the states is a symmetry, not an identification of microstates.

Let $w$ count the unequal neighbouring pairs, or domain walls. The energy is $E=-JL+Jw$. A single domain wall is impossible with periodic boundary conditions: following all the other, equal-state bonds around the ring would force its endpoints to be equal too. Two domain walls are possible, splitting the ring into two nonempty uniform arcs carrying different states. Thus the first excited state hasChoose the two wall bonds in $\left(\genfrac{}{}{0ex}{}{L}{2}\right)$ ways. For that choice, identify one of the resulting arcs by its smallest site label. Its state has $Q$ choices and the other arc has $Q-1$. This counts each configuration exactly once, giving the two-domain-wall degeneracy of a Potts ringFor example, $Q=2,L=4$ gives $d_0=2$ and $d_1=12$; the remaining two configurations are alternating and have four domain walls. The formula also covers $L=2$, where the specified ring energy counts the two directed neighbouring bonds, and an unequal pair breaks both. If $Q=1$ or $L=1$, every allowed configuration is a ground state and there is no first excited state. For $Q\ge 3$, the number of domain walls need not be even: three distinct states on a three-site ring give three, although one remains impossible.

The Boltzmann factor factorizes over bonds. Define the spin-chain transfer matrix byHere ${\delta }_{ab}$ is the Kronecker delta. Summing the product of these matrix elements over all configurations, and closing the final bond by periodic boundary conditions, gives the canonical partition functionHence, writing $I$ for the identity matrix and $P$ for the all-ones matrix,Every element of $P^2$ is a sum of $Q$ ones, while every diagonal element of $P$ is one. ThusMore explicitly, for $u=(1,\dots ,1{)}^T$, $Pu=Qu$. If ${\sum }_a{v}_a=0$, then $Pv=0$. These complementary eigenspaces have dimensions $1$ and $Q-1$. Equivalently, $P/Q$ is the orthogonal projector onto the linear span of $u$. The eigenvalues and their eigenvalue multiplicities are thereforeSince the two matrices share these eigenvectors, the exact finite-ring canonical partition function isBoth terms matter in the finite-ring low-temperature limit: discarding the second would lose the correct ground state degeneracy factor and introduce a spurious single-wall contribution.

Put $x=e^{-\beta J}$. Factoring out the ground state Boltzmann factor from the exact canonical partition function givesAt fixed finite $L$, the binomial theorem givesTaking the logarithm yields the finite-ring low-temperature expansion of the Helmholtz free energy,The constant term is the ground state energy. The term $-{\beta }^{-1}\log Q$ is the contribution of the ground state degeneracy factor. The first thermal correction follows directly by retaining the first excited state manifold:Each pair of domain walls costs $2J$, explaining the absence of an $e^{-\beta J}$ term. For $Q=2$, every configuration has an even number of domain walls, so the displayed remainder improves to $O_L(e^{-4\beta J}/\beta )$.

This is an expansion for fixed $L$, not a uniform thermodynamic limit. In particular, the small correction requires $\left(\genfrac{}{}{0ex}{}{L}{2}\right)(Q-1)x^2\ll 1$, with higher terms also controlled as $x\to 0$ at fixed $L,Q$. Taking $L\to \infty$ first instead selects ${\lambda }_{+}$ and givesIts leading correction is of order $x$. The finite-ring pair constraint and the extensive population of domain walls in an arbitrarily long chain explain the different ordering of these two limits.

Use $\hslash =1$ throughout. Let $N_a=a^{†}a$, $N_b=b^{†}b$ and $N=N_a+N_b$. Operators belonging to different boson species commute. The proposed Schwinger boson representation givesConsequently the first of the spin commutation relations isThe number operator commutators $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$ and their $b$ counterparts giveThus all the spin commutation relations hold. Also $[N,S^{\pm }]=[N,S^z]=0$: each fixed-$N$ sector is invariant. To identify its spin, calculate the Casimir operator:The irreducible spin representation with prescribed $s$ therefore requires $N=2s$, and its normalized Fock basis correspondence isConversely, $s=(n_a+n_b)/2$ and $m=(n_a-n_b)/2$. The nonnegative integer occupation numbers give precisely $m=-s,-s+1,\dots ,s$, and there are $2s+1$ such states. As a check, the spin raising operator acts aswhich is the usual spin ladder operator coefficient $\sqrt{s(s+1)-m(m+1)}$.

For a lattice of fixed spins, the local spin constraint in the Schwinger boson representation is essential:The full two-species bosonic Fock space also contains other local irreducible spin representations; fixing only the total boson number does not remove them.

For the specified positive-exponent discrete Fourier transform, the inverse discrete Fourier transforms have negative exponents:Every sum runs over $q=2\pi j/L$, $j=0,\dots ,L-1$. Indeed, orthogonality of roots of unity givesso substituting the forward discrete Fourier transform into either proposed inverse recovers the original operator. Taking Hermitian conjugates supplies the inverse formulas for the creation operators.

Multiplying the inverse discrete Fourier transforms gives the local number operator combinationThis local expression includes off-diagonal momentum terms. Only after summing over sites does orthogonality of roots of unity giveIn a physical fixed-$s$ state the first boxed operator acts as $s$ at every site, by the local spin constraint in the Schwinger boson representation.

Take $L\ge 2$, so neighbouring operators act at distinct sites. The identityexpresses the Hamiltonian operator of this quantum XY chain in spin ladder operators. Substituting the Schwinger boson representation gives the requested quartic expression:Operators at different sites commute, allowing normal ordering without a correction here. For the first quartic term, the inverse discrete Fourier transforms giveHere ${\delta }^{(L)}$ is one when its two momenta agree modulo $2\pi$ and zero otherwise; it expresses lattice momentum conservation. The other quartic term has the same ordered operators with phase $e^{i(p-r)}$. On the support of this Kronecker delta, $p-r=t-q$ modulo $2\pi$, so the two phases sum to $2\cos (q-t)$. Thus the Fourier vertex of a Schwinger boson chain isAn equivalent three-sum expression sets $t=p+q-r$ modulo $2\pi$, so its coefficient is $\cos (r-p)$. The factor $1/L$ comes from four factors $L^{-1/2}$ and one site sum, while the factor of two from the two exchange terms cancels the original $1/2$. The formula is Hermitian: conjugation exchanges $p$ with $t$ and $q$ with $r$, and the coefficient is unchanged after imposing lattice momentum conservation. It also preserves each local total boson number in its real-space form, hence preserves the physical constrained subspace.

Use $⟨O⟩=⟨G|O|G⟩$. The given state is a two-species uniform boson number state: only the $q=0$ mode of each species is occupied. Applying an annihilation operator to an unoccupied mode gives zero, while $A_0^{†}{A}_0$ has eigenvalue $n_A$. ThusThe two species have independent fixed occupation numbers, so mixed expectation values factorize. The inverse discrete Fourier transforms yieldIn the momentum-space Hamiltonian operator, only $p=q=r=t=0$ contributes: each annihilation operator must remove an occupied boson, and each creation operator must restore it to the same mode. HenceThere is no replacement of $n_B$ by $n_B+1$ in this energy: in the real-space calculation the relevant neighboring-site commutator is zero. Both expressions assume distinct neighbouring sites, as in ii.

To match the prescribed spin length in the mean, impose $n_A+n_B=2sL=:N$. At fixed integer $N$, the energy is minimized by maximizing $n_A{n}_B=[N^2-(n_A-n_B{)}^2]/4$. Consequently the best choices within this relaxed family areWhen $N$ is even, $n_A=n_B=sL$ and $⟨H⟩=-J{s}^{2}L$. When $N$ is odd, either assignment of $(N-1)/2,(N+1)/2$ gives $⟨H⟩=-J(N^2-1)/(4L)$. Without fixing $N$, this auxiliary boson Hamiltonian operator has no lower bound: taking $n_A=n_B\to \infty$ sends its expectation value to $-\infty$.

Next, $S_i^x=(a_i^{†}{b}_i+b_i^{†}{a}_i)/2$. Each term changes the two species' occupation numbers, producing a state orthogonal to $|G⟩$. ThereforeFor distinct sites $i,j$ modulo $L$, expand the product of the two spin operators. The terms $S_i^{+}{S}_j^{+}$ and $S_i^{-}{S}_j^{-}$ again have zero expectation value. The remaining terms giveHere $⟨b_i{b}_j^{†}⟩=n_B/L+{\delta }_{ij}$ and the Kronecker delta vanishes at distinct sites. ConsequentlyThe same holds for the $y$ components. At the balanced choice and as $L\to \infty$, each transverse two-point correlation function tends to $s^2/2$ independently of separation, whereas the one-point transverse magnetization is zero. This is transverse spin order in a number eigenstate: the state has a definite total $S^z=(n_A-n_B)/2$ and retains the global U(1) symmetry of rotations about $z$, yet has the transverse correlations of an ordered state with no selected common azimuth. The sum of the $x$ and $y$ correlations tends to $s^2$, describing ferromagnetic long-range order within this approximation. The thermodynamic limit is understood with both the separation and $L$ growing, without identifying the two sites periodically.

Fixed-spin qualification. The proposed state is not actually in the physical fixed-$s$ Hilbert space for $s>0,L>1$. It satisfies the local spin constraint in the Schwinger boson representation only in expectation value. For example, with $L=2,s=1/2,n_A=n_B=1$,The nonzero component $a_0^{†}{b}_0^{†}|\mathrm{vac}⟩/2$ has local total occupation numbers $(2,0)$ and hence local spins $(1,0)$, instead of $(1/2,1/2)$. More generally the local total number operator $N_n$ has varianceEach species contributes a binomial distribution at a given site, with success probability $1/L$, and the two contributions are independent. In the physical subspace that variance must be zero. The proposed state even hasby the Casimir operator identity from a. Thus the calculated energy is a relaxed mean-field theory estimate, not a Rayleigh-Ritz variational principle bound for the stated fixed-$s$ quantum XY chain. Projecting onto $N_n=2s$ at every site would give a physical trial state and generally change its expectation values; in the two-site example the projected state has energy $-J$, whereas the unprojected value is $-J/2$. The calculated long-range order characterizes the supplied approximation and does not establish long-range order in the exact one-dimensional ground state.